Consider a two-level atom of which the lower and upper levels are denoted, respectively, a and b. If spontaneous emission from the upper to the lower level is the only source of relaxation, then the rate of change of the diagonal density matrix terms will be:

$$ \frac{d \rho_{ee}}{dt} = - \Gamma_{sp} \rho_{ee} \qquad \frac{d \rho_{gg}}{dt} = \Gamma_{sp} \rho_{ee} $$

In most textbooks they then simply say that the relaxation of the coherence terms is given by

$$ \frac{d \rho_{eg}}{dt} = - \frac{\Gamma_{sp}}{2} \rho_{eg} $$

but without providing any explanation on why this is true. Is this supposed to be intuitive? Please explain me how I can derive this factor of $\frac{\Gamma_{sp}}{2}$ for the relaxation of coherence.


2 Answers 2


I do not know of a very intuitive way to derive this relation between the relaxation and decoherence rates that does not rest on other, potentially non-obvious assumptions (such as the form of the familiar Lindblad equation describing relaxation). But I will offer two arguments supporting it. The first may seem less rigorous since it relies on a particular model underlying the relaxation mechanism. However, it turns out the derived result is quite general and the fact that relaxation needs to be accompanied by some minimal amount of dephasing follows from a more general argument I present in the second part (and in fact, this minimal amount of dephasing is exactly the one we get from the simple relaxation model in the first part).

  1. We can model relaxation as an interaction of the two-level system $S$ with an environment $E$, followed by tracing out of $E$. In the simplest model, $E$ can be another two-level system which is initially prepared in its ground state $|0\rangle$. Evolving the composite system for a time $T$ under an interaction Hamiltonian $\hbar J(\sigma_{+}^{(S)}\sigma_{-}^{(E)} + \sigma_{-}^{(S)}\sigma_{+}^{(E)})$ leaves the state $|0\rangle_{S}|0\rangle_{E}$ unchanged and transforms $|1\rangle_{S}|0\rangle_{E}$ into $\cos\theta |1\rangle_{S}|0\rangle_{E} -\mathrm{i}\sin\theta|0\rangle_{S}|1\rangle_{E}$, where $\theta = JT$. Thus, when $S$ is initially in an arbitrary pure state $\alpha|0\rangle + \beta|1\rangle$, the composite system ends up in $(\alpha|0\rangle_{S}+\beta\cos\theta|1\rangle_{S})|0\rangle_{E} - \mathrm{i}\beta\sin\theta|0\rangle_{S}|1\rangle_{E}$ and after tracing out $E$, we get the density matrix $$ \left( \begin{array}{cc} |\alpha|^2 + |\beta|^2\sin^2\theta & \alpha\beta^{*}\cos\theta\\ \alpha^{*}\beta\cos\theta & |\beta|^2\cos^2\theta \end{array} \right). $$ Inspecting the diagonal elements, we see that a $\sin^2\theta$ fraction of the excited state population was transferred into the ground state. Therefore, this model will mimic the effect of the relaxation process described by the two evolution equations in the question over a time $T$ if we set $\cos^2\theta = \exp(-\Gamma_{sp}T)$. But when we look at the off-diagonal elements, we see they were reduced by a factor $\cos\theta = \exp(-\Gamma_{sp}T/2)$. So at least in this particular model, the factors $c_{01}$ and $c_{11}$ by which the off-diagonal density matrix elements and the excited state population are reduced, respectively, are always related by $c_{01} = \sqrt{c_{11}}$. If the evolution of the density matrix elements is to be described by a first-order differential equation (this is not obvious and relates to an assumption that the non-unitary evolution is Markovian), then the rate in the differential equations for the off-diagonal elements needs to be half the rate in the equation for the excited state population.

  2. We can also argue that the factors $c_{01}$ and $c_{11}$ need to satisfy $c_{01}\le \sqrt{c_{11}}$ (where equality corresponds to an ideal relaxation process, while $c_{01}$ can be lower when there is pure dephasing in addition) on very general grounds, assuming the relaxation process is a completely positive trace-preserving map (CPTP; a very general class of maps describing almost all commonly encountered cases of quantum mechanical evolution, whether it is unitary, non-unitary, involving measurements, feedback, etc.). This property means that the process maps any positive semidefinite density matrix onto another semidefinite matrix with the same trace. For a map of the form $$ \left( \begin{array}{cc} \rho_{00} & \rho_{01}\\ \rho_{10} & \rho_{11} \end{array} \right) \rightarrow \left( \begin{array}{cc} \rho_{00} + (1-c_{11})\rho_{11} & c_{01}\rho_{01}\\ c_{01}\rho_{10} & c_{11} \rho_{11} \end{array} \right) $$ to be CPTP, the final matrix needs to have non-negative trace and determinant (a property equivalent to semidefiniteness for $2\times 2$ matrices) for any initial semidefinite density matrix. The trace is non-negative automatically since the mapping we consider preserves it. We therefore only need $$ c_{11}\rho_{00}\rho_{11} + c_{11}(1-c_{11})\rho_{11}^2 - c_{01}^2 \rho_{01}\rho_{10} \ge 0. $$ Any initial semidefinite density matrix can be written in the form $(1+\vec{a}\cdot\vec{\sigma})/2$ with the Bloch vector $\vec{a}$ satisfying $|\vec{a}|\le 1$. In terms of $\vec{a}$, our condition for $c_{01}$ and $c_{11}$ becomes $$ c_{11}(1-a_z^2) + c_{11}(1-c_{11})(1-a_z)^2 - c_{01}^2(a_x^2 + a_y^2) \ge 0 $$ for any $\vec{a}$ such that $|\vec{a}|\le 1$. Therefore \begin{align*} c_{01}^2 \le \inf_{|\vec{a}|\le 1}\frac{ c_{11}(1-a_z^2) + c_{11}(1-c_{11})(1-a_z)^2}{ a_x^2 + a_y^2 } &= \inf_{0\le |a_z|\le a\le 1}\frac{ c_{11}(1-a_z^2) + c_{11}(1-c_{11})(1-a_z)^2}{ a^2 - a_z^2 }\\ &= \inf_{0\le |a_z|\le 1}\frac{ c_{11}(1-a_z^2) + c_{11}(1-c_{11})(1-a_z)^2}{ 1 - a_z^2 }\\ &= c_{11} + c_{11}(1-c_{11})\inf_{0\le |a_z|\le 1}\frac{ 1-a_z}{ 1+a_z }\\ &= c_{11}, \end{align*} exactly as I claimed.


This fact can be derived from the Lindbald master equation:

$$ \frac{\partial \rho}{\partial t} = \frac{1}{i \hbar} [H, \rho] + L \rho L^{+} - \frac{1}{2}\{L^+L,\rho \} $$

with L being the "jump operator". In the case of spontaneous emission in a two level system $$L = \sqrt{\Gamma}\sigma_-$$

A "simple" derivation of the Lindbald master equation can be found here. It is not short, but well explained, and I couldn't find better "intuitive" motivations for this equation.

If there is no external interaction, and only spontaneous emission, then $[H,\rho] = 0$.

By putting $$ \rho = \begin{pmatrix} \rho_{ee} & \rho_{eg}\\ \rho_{ge} & \rho_{gg} \end{pmatrix} $$

inside this formula we get:

$$ \begin{pmatrix} \dot{\rho_{ee}} & \dot{\rho_{eg}}\\ \dot{\rho_{ge}} & \dot{\rho_{gg}}\\ \end{pmatrix} = \Gamma \cdot \begin{pmatrix} 0 & 0\\ 0 & \rho_{ee}\\ \end{pmatrix} - \frac{\Gamma}{2} \cdot \begin{pmatrix} 2\rho_{ee} & \rho_{eg}\\ \rho_{ge} & 0\\ \end{pmatrix} = \Gamma \cdot \begin{pmatrix} -\rho_{ee} & -\frac{1}{2}\rho_{eg}\\ -\frac{1}{2}\rho_{ge} & \rho_{ee}\\ \end{pmatrix} $$

This shows indeed that the relaxation of coherence rate (the terms in the anti-diagonal) are half the spontaneous emission rate (the terms in the diagonal).


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